狭窄框架下保险模型的Stackelberg博弈问题.pdf

Advances in Applied Mathematics A?,2024,13(3),912-927Published Online March 2024 in Hans.https:/www.hanspub.org/journal/aamhttps:/doi.org/10.12677/aam.2024.133086dee?x?.?StackelbergK?&?n?U9vF2024c2?13FF2024c3?8FuF2024c3?14F de-?OKe?Stackelberg 2?xKdeX?x?LxkU?x?wxd?g?5?x?”?=de3 Stackelberg 2?xik?xiJn?5?,?xi?KJ?kb?2?xi?K?xiLJ?5zL?-?x?g,?2?xiLJ?zL?-?d?,?Stackelberg?Ku?xi5c?8I(I)=E(PI)?K?x?L?x?z2?xiL?OKO?dr?L Taylor m?)?CqLcStackelberg-?OKde2?x?OKA Stackelberg Game Problem inInsurance Models with Narrow FramingShaodi SunSchool of Science,Hebei University of Technology,Tianjin:?&.dee?x?.?StackelbergKJ.A?,2024,13(3):912-927.DOI:10.12677/aam.2024.133086?&Received:Feb.13th,2024;accepted:Mar.8th,2024;published:Mar.14th,2024AbstractIn this paper,we consider the Stackelberg game reinsurance problem under mean-variance criterion with narrow framing.The motivation for purchasing insurancemight not only be hedging wealth risk,but also to consider the purchase of insuranceitself as a risky investment,which is called narrow framing.Inspired by this,we usea quadratic utility function to measure the local gain-loss utility of the net benefits ofinsurance,namely,narrow framing.As the Stackelberg game in insurance models,thereinsurer first offers the insurer a reasonable indemnity in exchange for the appropri-ate premium.Then,the insurer selects an indemnity based on the premium principle.In this paper,suppose that the reinsurer chooses the expected value premium prin-ciple,we compute the optimal insurance indemnity to maximize the mean-variancefunctional of the insurers terminal wealth and the quadratic function of net insurancereturns.Then,given the optimal indemnity of the insurer,we compute the parameterof the expected value premium principle to maximize the mean-variance function ofthe reinsurers terminal wealth.In addition,we consider another Stackelberg game.For the insurer,by considering the same objective function as the former and givingthe premium principle(I)=E(PI),we obtain an expression of the optimal indem-nity.Afterwards,given the optimal indemnity,we maximize the expected utility forthe terminal wealth of the reinsurer and compute the optimal price intensity of thepremium.Furthermore,by applying Taylor expansion,we find the approximationexpression of the optimal solution.KeywordsStackelberg Game,Mean-Variance Criterion,Narrow Framing,Reinsurance,Expected Utility CriterionCopyright c?2024 by author(s)and Hans Publishers Inc.This work is licensed under the Creative Commons Attribution International License(CC BY 4.0).http:/creativecommons.org/licenses/by/4.0/DOI:10.12677/aam.2024.133086913A?&1.u2?x?O?mM5J?Borch 1L?z?xi?ox?L3?Ke2?x?K=5?x?O?zKaluszka 2-?KO“?K Borch 1?(J?1?2Young 3 Kaluszka 4O Wangs?K?K?2?x,?Borch 5?xik?2?xU?2?xi?d3?x?K?xV?|-?xdLp?xig3?x5?2?|,XJ?xd$2?xi2mCai?6l?xV?2?x?O32?x?O?8ICAc?Stackelberg?.52?x?Kz 7,80?k Stackelberg?SNChan Gerber 9Lz?xi2?xiL?Bowley)3?.?g.5O?2?xd3”7?Cq)3?Ac?Bowley)Y3x+ne?-Cheung?103?K-xe*?Chan Gerber 9?zLk(?xi?G?(?2?xi?dChi?11|Bowley)5)?xi2?xim?x=K?.?1?2?x?N?2?xi?x 1?(?2?xi|dz?Boonen?12?&Ee?Bowley)b?2?xi?)?xi?&ELi Young 133L?zIOe?Stackelberg?Bowley)k-?K5z?x?5O?)?x3?JK3?x?X/?xx?xDLb?k?U?n5?y?Xd?J?l?n513?x?uy?n)?x?L?k3 Heaton Lucas 14=?xxNvk?L?Curcuru?15?L?X?z?%?NLx?9u|?z?Barberis Huang 16?/?x?L?deX“n?u?(J?=m?/LoL?zz 17,18(JLdek?3 A?vku?n?de?5Barberis?19deB 5?l?|?|?dBarberis Huang 16Jl(Jm?/?Gottlieb Mitchell 20J?xde?.LdeK?on?x?U5?$uYKx?K?.Zheng 21?de?x?xb?3?xk*:Lx?xi?(JLde?xIku)|*?$?DOI:10.12677/aam.2024.133086914A?&xIk?L?xBehaghel Blau 22de”?)3 65?-O.3Chi?23 S/deL?.?x?L?9?x?L?xJK?N3?:?Stackelberg=+?-2?xik1?xi?1y?2?x3 Stackelberg L)fK5)2?xK3 zvkden Stackelberg(d?3?xi?8IKde?KN/3?1SN1fK?Kz?xiL?-?OK9?x?g?x?.1?fKz2?xiL?-?OKO?K?.31?SN1SN?/O(I)=E(PI)z2?xiL?OK?x?dr?)?eSNSXe1 2!?.b?1 3!?K3-?OKe Stackelberg K?1)O?xi?y?)?1 4!?/92?xi?8IOK?xi?9?dr?)|Taylor m?)?Cq?1 5!?1o(2.?.b?b?|k t=0,1t=0 L?xi3 t=1”Xr”X 3m 0,wi wi?xi?Lb?SX(x)L X?)FX(x)L X?2?x(I,(I)I 2?x?k?x 0k 0 I(x)x(I)2?x?b?R?xi?2?x?”k R=X IIS?32?x?K?z-?OK?n“n?Lu Stackelberg e?O?2?x5?xi2?xim?pK3d:de?Vg=b?2?x?xiw2?xi?X?xi?u(J?3?Y?g?5TK?Stackelberg K31SNXeb?(1)b?xi2?xi?LO wi wrLO Wi WrvXeXWi=wi X+I (I),(1)Wr=wr I+(I).(2)3?x?xi?x I(I)b?xi?g?xDOI:10.12677/aam.2024.133086915A?&?g g()?LXeg(x)=x cx2,(3)c v 0 c 0(3)?xi?8IJ?x?IzL?x?maxIEWi12V ar(Wi)+kE(g(I (I),(4)1 0?xi?”?Xk 0 de5?xi?x?3?-?k=0?xi5L?-?OK?k +?x=w2?xi?)I2?xiJ?zL?-?maxEWr22V ar(Wr),(5)2 0 2?xi?”?X3?1?SNeb?(1)3TSN?K5)Stackelberg?)b?/(I)=E(PI),C P Ldrv P 0 EP=1(2)?dr P?xi?8IJ?x?I5zL?x?maxIEWi12V ar(Wi)+kE(g(I (I),?)I2?xiLzL?)?2?xdrmaxPEWr.(6)1?y?x=I (I)g 4O?DOI:10.12677/aam.2024.133086916A?&3.1?.)?!O Stackelberg?)L1?Kz?xiL?-?9?x?gO?x?wL1?z2?xi?-?)2?xi?eu?1)3.1.?xiK9)n1.z?xi?8I?I=I(;)vXefI(x;)=(1x )+1+2ck x,(7)0 e(8)?)(1+k)=1Z10SX(x)dx 2ck121+2ckZ1SX(x)dx.(8)y km?xi)?8I EWi12V ar(Wi)+kE(g(I (I)Wid(1)g(I (I)d(3)8IzmaxIwi EX 12E(X2)+12(EX)2(1+k)EI?12+ck?E(I2)+?12+ck(1 2)?(EI)2+1E(XI)1EXEI.(9)?(9)?d f(I)Lf(I)=(1+k)+1EXEI?12+ck?E(I2)+12+ck(1 2)(EI)2+1E(XI).(10)?z fb?EI=m 0,=EX?f(I)=(12+ck)E(I2)+1E(XI)?K=z)f(I)?)?zK.KF#?z?1).KFf R?#L(I)LL(I)=Z0?12+ck?I2(x)+1xI(x)I(x)?dFX(x)+m.I(x)?ILI(x;)=(1x )+1+2ck x.DOI:10.12677/aam.2024.133086917A?&m?L?m=E?(1X )+1+2ck X?.(11)?e5y m 0,3?R?(11)?YOB(11)?Xe/m=R2ck0SX(x)dx+11+2ckR2ckSX(x)dx,0,11+2ckR1SX(x)dx,0.?m0=(11+2ckSX(2ck),0,11+2ckSX(1),0.(12)l(12)?l O?m l?0 m AX?)dz(4)5 I(x)?K=z m?5z(10)?foL 5O I(x)f?u?f()=?(1+k)+1EX?g1()?12+ck?g2()+?12+ck(1 2)?(g1()2+1g3(),g1()=EI=m,g2()=E(I2)=2R2ck0 xSX(x)dx+21(1+2ck)2R2ck(1x )SX(x)dx,0,21(1+2ck)2R1(1x )SX(x)dx,0,g3()=E(XI)=2R2ck0 xSX(x)dx+11+2ckR2ck(21x )SX(x)dx,0,11+2ckR1(21x )SX(x)dx,0.O g1g2 g3?g01()=m0=(11+2ckSX(2ck),0,11+2ckSX(1),0,g02()=ck(1+2ck)SX(2ck)21(1+2ck)2R2ckSX(x)dx,0,21(1+2ck)2R1SX(x)dx,0,g03()=2ck(1+2ck)SX(2ck)11+2ckR2ckSX(x)dx,0,1(1+2ck)SX(1)11+2ckR1SX(x)dx,0.DOI:10.12677/aam.2024.133086918A?&?0?kf0()=?(1+k)+1EX?g01()(12+ck)g02()+1+2ck(1 2)g1()g01()+1g03()=SX(2ck)1+2ck(1+k)2ck(1 2)Z2ck0SX(x)dx+2ck121+2ckZ2ckSX(x)dx!.?l O?0)S?l(1+k)+2ck121+2ck O?d 0okf0()0?0 f0()=?(1+k)+1EX?g01()?12+ck?g02()+?1+2ck(1 2)?g1()g01()+1g03()=SX(1)1+2ck(1+k)+1Z10SX(x)dx+2ck121+2ckZ1SX(x)dx!.l?Lw?l 0 O?)S?kO?o3?f()?ve(1+k)=1Z10SX(x)dx 2ck121+2ckZ1SX(x)dx.n2.?I=I(;d)?LI(x;d)=(x d)+,(13)d=1 0 e?)(1+k)=1Zd0FX(x)dx 2ck2ZdSX(x)dx,(14)k=11+2ck.(15)y (7)I(x;d)?Lz(13)?/O?(15)?9 d=151 l(13)?L/w?”2?x)?d 2?x.?xi?x?XO?d$?2?x?O52 deX2?x?Kw?dO?xi?k Cu?xL?x?w2?xi?xi2?x)?x|?$?DOI:10.12677/aam.2024.133086919A?&3.2.2?xiK9)?)Stackelberg?,fK2?xi?K(5)?8I EWr22V ar(Wr)m Wrd(2)EWr22V ar(Wr)=wr+E(X d)+222V ar(X d)+.(16)dz(16)?duz g(,d)g(,d)=E(X d)+22V ar(X d)+.(17)(14)?LO=p(1+k)2+8c1kE(X d)+(d E(X d)(1+k)4ckE(X d)+.d(17)-#?g(d)=p(1+k)2+8c1kE(X d)+(d E(X d)(1+k)4ck22V ar(X d)+.g(d)?g0(d)=1(E(X d)d)SX(d)+E(X d)+(1 SX(d)p(1+k)2+8c1kE(X d)+(d E(X d)+2E(X d)+(1 SX(d).(18)d?d(18)?)?!ln?ef?d?13.3 O?n?1?!Y.3?!?Jwi=10,c=0.01,1=0.25,2=0.2,k=3.4,=11+2ck=0.786.(1)b?”C X l?K)LSx(x)=(ex,x (0,10,1 e10,x=0.)?LOeE(X d)d=Zd0SX(x)dx d=1 ed d,E(X d)+=Z10dSX(x)dx=ed e10.DOI:10.12677/aam.2024.133086920A?&d“(18)g0(d)=1?(1 ed d)ed+(ed e10)(1 ed)?p(1+k)2+8c1k(ed e10)(d+ed 1)+2?ed e10?1 ed?.de?)0.25?(1 ed d)ed+(ed e10)(1 ed)?p19.36+0.0534(ed e10)(1 ed)+0.1572?ed e10?1 ed?=0.(19)(19)?d=4.70.(2)b?”C X l?Pareto)SX(x)=?1211201(x+1)21120?Ix0,10.kOE(X d)d=Zd0SX(x)dx d=121120dd+1121120d,E(X d)+=Z10dSX(x)dx=1211201d+1+d12021120.O(J?g0(d)=0.25h(dd+1 d)(1211201(d+1)21120)+(1211201d+1+d12021120)(1 1(d+1)2)iq19.36+0.0534(1211201d+1+d12021120)(121120d 121120dd+1)+0.1572?1211201d+1+d12021120?1 1(d+1)2?.-g0(d)=0O?d=6.314.1?.)?!O1?SN?Stackelberg)1)?K?drC P?xi?8I1SN?8I=zL?-?9?x?gO?x?1?z2?xi?)2?xi?dreu?1)n3.3-?Ke?xi?K?drC Pz?xDOI:10.12677/aam.2024.133086921A?&i?8I?xi?x?)I=q1X+q2P+n=12(1+2ck)X+12(1+2ck)ckq212Xq3(1+k)+n.(20)3?zOKe2?xi?8IO?dr?)P=PX(X )+1=q3(1+k)2ckq12X(X )+1,(21)q1,q2 q3?LOd(23)(24)(27)XO“L”X?IO?y?xi?8IzL?-?x?g?Lm EWi12V ar(Wi)+kE(g(I (I)Wi g(I (I)d(1)(3)?xi?8I?dmaxI(1+k)EI (1+k)E(PI)(12+ck)E(I2)+12(EI)2+1E(XI)1EXEI ckE(PI)2+2ckEIE(PI).b?I z8IC Q?It=I+tQ t L?g(t)g(t)=(1+k)E(It)(1+k)E(PIt)(12+ck)E(I2t)+12E(It)2+1E(XIt)1EXE(It)ckE(PIt)2+2ckE(It)E(PIt).(22)g(t)3 t=0?=g0(0)=0(22)u t?g0(t)g0(t)=(1+k)EQ (1+k)E(PQ)(1+2ck)E(QIt)+1E(It)EQ+1E(XQ)1EXEQ 2ckE(PIt)E(PQ)+2ckEQE(PIt)+2ckE(It)E(PQ).X g0(0)=0?eEQ(1+k)(1+k)P (1+2ck)I+1EI+1X 1EX2ckPE(PI)+2ckE(PI)+2ckPEI=0.du Q?C)S?0z?I=q1X+q2P+n,DOI:10.12677/aam.2024.133086922A?&n?q1 q2?Lq1=11+2ck,(23)q2=2ckq1 (1+k)2ckq1E(PX)1+2ckE(P2).(24)?e5?2?xi?K2?xiL?z(6)?d maxPEI(I)I?L“T8I2?xi?8I?d?zmaxPq1EX+q1E(PX)+q2E(P2)q2.(25)(24)q2?L?eXq1EX q1E(PX)=(1+2ckE(P2)q2+k+12ck.“?(25)8I?dmaxP(1+12ck)q21+k2ck.d2?xi?8I?d?z q2min,Pq2=min,P 2ckq1XP(1+k)2ck2P+2ck+1,C X P?XX PO X P?IO?e?zT8Iw,k =1.d8I?dmaxP2ckq1XP+k+12ck2P+2ck+1.8I?O P?P=p(1+k)2+2ckq21(2ck+i)2X(1+k)2ckq1X.(26)duX =1b?P X vX P=aX+b(eE(P)=1,V ar(P)=2P.DOI:10.12677/aam.2024.133086923A?&OX a,b?a=PX,b=1 PX.o?P?)P=PX(X )+1=1+q3(1+k)2ckq12X(X ),q3?Lq3=q(1+k)2+2ckq21(2ck+1)2X.(27)dz(24)q2?L-#?q2=2ckq1XP+k+12ck2P+1+2ck.(26)P?L q2?q2=ckq212Xq3(1+k).d?xi?I?LI=12(1+2ck)X+12(1+2ck)ckq212Xq3(1+k)+n.n4.3C X?IO?Xv?e Taylor mO?x?dr?Cq)I12(1+2ck)X+12(1+2ck)1+k1+2ck+n,P12(1+k)(X )+1.y b?Xv?Taylor m PCqP1X2(1+k),?dr P?CqP 1+12(1+k)(X ).DOI:10.12677/aam.2024.133086924A?&dq2Cqq2 1+k1+2ck,?I?CqI12(1+2ck)X+12(1+2ck)1+k1+2ck+n.n5.?dr?L2?xi?E(Wr)=wr+?q3(1+k)4ck2ckq21q3(1+k)?2X.?xi?cL(E(Wi),V ar(Wi)=?wi +?2ckq21q3(1+k)q3(1+k)4ck?2X,?2ck1+2ck+214(1+2ck)2?2X?,q3d(27).y (20)I?L(21)P?LO?2?xi?E(Wr)=wr EI+(I)=wr+?q3(1+k)4ck2ckq21q3(1+k)?2X.O E(Wi)Wid(1)?xiL?AE(Wi)=wi +?2ckq21q3(1+k)q3(1+k)4ck?2X.Lm V ar(Wi)z 2X+E(I)2(EI)2 2E(XI)+2EIO?E(I)2(EI)2=212X4(1+2ck)2,?xiL?A?V ar(Wi)=?2ck1+2ck+214(1+2ck)2?2X.DOI:10.12677/aam.2024.133086925A?&5.(?3?xi2?xiV?2?x?Kuden?Stackelberg?.?)31SNkz?xiL?-?9?x?g?xi?wLy?)?5lLw?x?d?x?,?z2?xiL?-?OK?v?31?SN(I)=E(PI)?/O?2?dr?)?L Taylor m?2?dr?Cq)z1 Borch,K.H.(1960)An Attempt to Determine the Optimum Amount of Stop Loss 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